#Stax, 26 24 18 bytes

:u{m*_{y#m:u_hy#h* 

Try it online!

Shortest solution so far that only uses printable ASCIIs Beaten by MATL.

Guess I was approaching the problem the wrong way. Repeating a working block is neither golfy nor interesting. Now at least it looks better ...

Explanation

:u{m* produces some garbage that does not affect the output.

_{y#m:u_hy#h* _{y#m map each character to its number of occurences in the string :u all counts are equal (result 1) _hy# get the count of appearance for the first character h halve it and take the floor, so that 1 becomes 0(result 2) * multiply the two results 

Stax, 26 24 18 bytes

:u{m*_{y#m:u_hy#h* 

Try it online!

Shortest solution so far that only uses printable ASCIIs Beaten by MATL.

Guess I was approaching the problem the wrong way. Repeating a working block is neither golfy nor interesting. Now at least it looks better ...

Explanation

:u{m* produces some garbage that does not affect the output.

_{y#m:u_hy#h* _{y#m map each character to its number of occurences in the string :u all counts are equal (result 1) _hy# get the count of appearance for the first character h halve it and take the floor, so that 1 becomes 0(result 2) * multiply the two results 

#Stax, 26 24 18 bytes

:u{m*_{y#m:u_hy#h* 

Try it online!

Shortest solution so far that only uses printable ASCIIs (or not?)

Guess I was approaching the problem the wrong way. Repeating a working block is neither golfy nor interesting. Now at least it looks better ...

Explanation

:u{m* produces some garbage that does not affect the output.

_{y#m:u_hy#h* _{y#m map each character to its number of occurences in the string :u all counts are equal (result 1) _hy# get the count of appearance for the first character h halve it and take the floor, so that 1 becomes 0(result 2) * multiply the two results 

#Stax, 26 24 18 bytes

:u{m*_{y#m:u_hy#h* 

Try it online!

Shortest solution so far that only uses printable ASCIIs Beaten by MATL.

Guess I was approaching the problem the wrong way. Repeating a working block is neither golfy nor interesting. Now at least it looks better ...

Explanation

:u{m* produces some garbage that does not affect the output.

_{y#m:u_hy#h* _{y#m map each character to its number of occurences in the string :u all counts are equal (result 1) _hy# get the count of appearance for the first character h halve it and take the floor, so that 1 becomes 0(result 2) * multiply the two results 

#Stax, 26 24 18 bytes

:u{m*_{y#m:u_hy#h* 

Try it online!

Shortest solution so far that only uses printable ASCIIs (or not?)

Guess I was approaching the problem the wrong way. Repeating a working block is neither golfy nor interesting. Now at least it looks better ...

Explanation

:u{m* produces some garbage that does not affect the output.

_{y#m:u_hy#h* _{y#m map each character to its number of occurences in the string :u all counts are equal (result 1) _hy# get the count of appearance for the first character h half it and take the floor, so that 1 becomes 0(result 2) * multiply the two results 

#Stax, 26 24 18 bytes

:u{m*_{y#m:u_hy#h* 

Try it online!

Shortest solution so far that only uses printable ASCIIs (or not?)

Guess I was approaching the problem the wrong way. Repeating a working block is neither golfy nor interesting. Now at least it looks better ...

Explanation

:u{m* produces some garbage that does not affect the output.

_{y#m:u_hy#h* _{y#m map each character to its number of occurences in the string :u all counts are equal (result 1) _hy# get the count of appearance for the first character h halve it and take the floor, so that 1 becomes 0(result 2) * multiply the two results